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Let F be a field and let K be an algebraically closed field with F Ã¢?? K.

If f Ã¢??F[x] is irreducible (i.e. if f = m * n , then one of m or n is a constant) and f

has a multiple zero in K ,

then f Ã¢?Â² = 0

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Solution Preview

Proposition. Let F be a field and let K be an algebraically closed field with FâŠ†K. If fâˆˆF[x] is irreducible and f has a multiple zero in K, then f^'=0.

Proof: Before we prove this proposition, let's introduce a theorem related to separable polynomials and their derivatives. A polynomial fâˆˆF[x] is said to be separable if it has distinct zeros in a splitting field over F. That is, each zero of f has multiplicity 1. We now introduce the following theorem.

Theorem 1. A nonzero polynomial f in F[x] is separable if and only if it's relatively ...

Solution Summary

Proposition is clarified here for an algebraically closed field.

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